One question about this theorem:
Suppose $A$ is a Banach Algebra, $x \in G(A)$, $h \in A$ and $\left\lVert h \right\rVert < \frac{1}{2} \left\lVert x^{-1} \right\rVert^{-1}$. Then $x + h \in G(A)$ and
$$ \left\lVert (x + h)^{-1} - x^{-1} + x^{-1}hx^{-1} \right\rVert \leq 2 \left\lVert x^{-1} \right\rVert^3 \left\lVert h \right\rVert^2 $$
Proof : Since $x + h = x(e + x^{-1}h)$ and $\left\lVert x^{-1}h \right\rVert < \frac{1}{2}$, theorem 10.7 implies that $x + h \in G(A)$ and that the norm of the right member of the identity $$ (x + h)^{-1} - x^{-1} + x^{-1}hx^{-1} = \left[ (e + hx^{-1})^{-1} - e + x^{-1}hx \right]x^{-1} $$ is at most $2\left\lVert x^{-1} h \right\rVert^2 \left\lVert x^{-1} \right\rVert$
Now the doubt I have is that the statement of the theorem is wrong because the statement gives the bound $2\left\lVert x^{-1} \right\rVert^3 \left\lVert h \right\rVert^2$ but the proved one is $2\left\lVert x^{-1} h \right\rVert^2 \left\lVert x^{-1} \right\rVert$,
am I missing anything?
From $10.7$, given an invertible element $x$ such that $\|x\| < 1$,
$$ \|(e-x)^{-1} - e - x\| \le \frac{\|x\|^2}{1-\|x\|}.$$
In our case (typo aside), we have
$$ \|\big((e+x^{-1}h)^{-1} - e + x^{-1}h\big)x^{-1}\| \le \|(e+x^{-1}h)^{-1} - e + x^{-1}h\| \|x^{-1}\|. $$
By $10.7$, letting $x\mapsto -x^{-1}h$, we see that
$$ \|\big((e+x^{-1}h)^{-1} - e + x^{-1}h\big)x^{-1}\| \le \frac{\|x^{-1}h\|^2}{1-\|x^{-1}h\|} \|x^{-1}\| \le \frac{\|x^{-1}\|^3 \|h\|^2}{1-\|x^{-1}h\|}. $$
From the assumption that $\|h\| < \frac{1}{2}\|x^{-1}\|^{-1}$, we have that $\|x^{-1}h\| \le \|x^{-1}\|\|h\| < \frac{1}{2}$ so that $1-\|x^{-1}h\| \ge 1-\frac{1}{2} = \frac{1}{2}$. Taking a reciprocal gives
$$ \frac{1}{1-\|x^{-1}h\|} \le 2. $$
Piecing this together, we have
$$\|\big((e+x^{-1}h)^{-1} - e + x^{-1}h\big)x^{-1}\| \le 2 \|x^{-1}\|^3\|h\|^2.$$
I suspect this was a typo by Rudin or perhaps something that got lost in the editing process. It is a pretty substantial book, after all.